According to Heath, this proof is really due to Archimedes. It should be mentioned that it is of course a lot easier to prove the result using trigonometry! The area of a triangle is 1/2ab.sinC, and using c2=a2+b2-2ab.cosC, and sin2+cos2=1, the result follows in a few lines.

I think the following argument also constitutes a proof: consider the square of the area of a triangle of sides a, b, c. Dimensionally, it must be a homogeneous expression of fourth degree (a quartic) in a, b, c. It must also be symmetric in a, b, c (it can't matter which sides of a given triangle are labelled with which letters), and the area of the triangle must be zero if the length of any one of the sides equals the sum of the other two, -- so a+b-c must be a factor, etc. This gives Heron's expression, except for an overall multiplicative factor which can be determined by taking an easy example, such as a triangle of sides 1, 1, sqrt2. (Obviously, there is no quadratic expression in a, b, c satisfying our requirements, so we are forced to consider the square of the area rather than the area itself.)